Probability: Draw your own card In the anime HunterXHunter, there is an episode where 24 people are assigned a numbered ID tag, and each person draws a card with one of those numbers on it. Each person is supposed to hunt the person who's number is on the card they draw and steal their ID tag. A friend and I were discussing this and trying to determine the probability of one of those people drawing his or her own number, but we came up with two different answers.
She says that the probability is 1/24, since one of the 24 cards has that person's own number. Multiply that by 24 hunters, and you therefor have basically a 100% chance of someone drawing their own number each time the game is played, regardless of how many hunters there are.
That doesn't feel right to me, and I think that the actual probability is 1/576, or 1/24^2. My reasoning being that each hunter is assigned a number at random, and then they choose another number at random when they draw a card, so they ultimately take two consecutive 1/24 chances. Multiply that probability by each hunter and you have a 24/576, or approximately 1/27, chance that someone will draw their own number. Since the equation becomes x/x^2, the probability of this happening decreases exponentially the more hunters there are in the game.
These numbers make the most sense to me, but my friend insists that the probabilistic assignment of each hunter's number is irrelevant because they're all unique, like names, which I can't entirely argue against. Can someone break our stalemate by explaining which of us, if either, is correct? What is the probability of one hunter drawing his or her number in this manner?
 A: See the wikipedia article on derangements.  1-the probability that no people draws his or her own ID number = at least some one gets his or her own ID number and by the above article this is almost $1-\dfrac{1}{e}$
A: You're correct that the probability of a single person drawing their-own card is 1/24. But the probability that at least one person draws their own card isn't 1. The correct way to compute it is to say that the probability that no-one will draw their-own card is (1-1/24)^24, therefore the probability that at least one person will draw their-own card is 1 - (1-1/24)^24, or about 64%.
A: Voldemort is correct, the article on derangements will help (if you can understand it). There is no simple, exact formula that I know of. The correct answer is that it is one minus the number of derangements of 24 elements divided by the number of possible permutations of 24 elements. So 1-(!24/24!). Jim Peters correctly explains that 1/24 is the probability that a particular person will not get their own ID back, but you want the probability that at least one person gets theirs back. But this is not just (1/24)^24 since each person getting or not getting their card back is not independent of the others since they draw without replacement. This is easy to see if there were only two people say Alice and Bob. Then the probability that Alice gets hers is 1/2 and the probability that Bob gets his is also 1/2. But they aren't independent since they get each others in the same cases. With two people, the probability that anyone gets theirs back is 1/2, not (1/2)^2. 
